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PS - perimeters/area

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PS - perimeters/area

by Xbond » Sun Nov 01, 2009 1:28 pm
Hi there,

I dont' know how to attack this PS ? Could you help me to understand the concept tested ?

The perimeters of square region S and rectangular region R are equal. If the sides of R are in the ratio 2:3.
What is the ratio of area of region R to area of region S ?
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by gmatv09 » Sun Nov 01, 2009 2:46 pm
Let a be the side of the square
l and w be the sides of the rectangle.

According to the problem:

4a = 2l + 2w
or a = (l + w)/2 --- (1)

l/w = 2/3 ---- (2)

a^2/lw = ?

We know that side of square = avg of sides of rect.
and sides of rect are in the ratio = 2/3

Choose smart nos. for the sides of rect = 4 * 6

Ans. 25/24
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by ayushr » Sun Nov 01, 2009 7:37 pm
gmatv09 wrote:
a^2/lw = ?

Ans. 25/24

What is the ratio of area of region R to area of region S ?
We need to find - lw/a^2

Answer should be - 25/24

Another way to solve:
Sides of R can be written as 2x and 3x
Preimeter of R = 2x(2x + 3x)= 10x
Area = 6x^2

Perimeter of Square, S = 4s = 10x
Side of S = 10x/4= 5x/2
Area= 25x^2/4

Area of R / Area of S = 24/25
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